Question

Find the partial derivative of the function f(x,y)=Integral of cos(-7t^2-6t-1)dt. Find fx(x,y) and fy(x,y)

Answer 1

`\mathbf{(\partial f)/(\partial x)= cos (-7x^2 -6x - 1)}`

`\mathbf{(\partial f)/(\partial y)= cos ( -7y^2 -6y-1)}`

Explanation:

Given that :

`f(x,y) = \int ^x_y cos (-7t^2 -6t-1) dt`

Using the Leibnitz rule of differentiation,

`(d)/(dt) \int ^(b(t))_(a(t)) f(x,t) dt= f(b(t),t) *b'(t) -f(a(t),t) * a' (t) + \int^(b(t))_(a(t)) (\partial f)/(\partial t) \ dt`

To find: fx(x,y)

`(\partial f)/(\partial x)= (\partial )/(\partial x) [ \int ^x_y cos (-7t^2 -6t -1 ) \ dt]`

`(\partial f)/(\partial x)= (\partial x)/(\partial x) cos (-7x^2 -6x -1 ) - (\partial y)/(\partial x) * cos (-7y^2 -6y-1) + \int ^x_y [(\partial )/(\partial x) \ \{cos (-7t^2-6t-1)\}] \ dt`

`(\partial f)/(\partial x)= cos (-7x^2 -6x - 1) -0+0`

`\mathbf{(\partial f)/(\partial x)= cos (-7x^2 -6x - 1)}`

To find: fy(x,y)

`(\partial f)/(\partial y)= (\partial )/(\partial y) [ \int ^x_y cos (-7t^2 -6t -1 ) \ dt]`

`(\partial f)/(\partial y)= (\partial x)/(\partial y) cos (-7x^2 -6x -1 ) - (\partial y)/(\partial y) * cos (-7y^2 -6y-1) + \int ^x_y [(\partial )/(\partial y) \ \{cos (-7t^2-6t-1)\}] \ dt`

`(\partial f)/(\partial y)= 0 - cos ( -7y^2 -6y-1)+0`

`\mathbf{(\partial f)/(\partial y)= cos ( -7y^2 -6y-1)}`